通过移除给定的 N 张牌中的最后一张，找出赢得游戏的玩家

给定两个整数N和K ，其中N表示游戏开始时出现的牌总数， K表示单回合可以移除的最大牌数。两名玩家A和B轮流取出最多K张牌，从玩家A开始一张一张。取出最后一张牌的玩家为赢家。任务是检查A是否可以赢得比赛。如果发现是真的，打印“A”作为答案。否则，打印‘B’ 。

例子：

Input: N = 14, K = 10
Output: Yes
Explanation: 
Turn 1: A removes 3 cards in his first turn. 
Turn 2: B removes any number of cards from the range [1 – 10] 
Finally, A can remove all remaining cards and wins the game, as the number of remaining cards after turn 2 will be ≤ 10

Input: N = 11, K=10
Output: No

方法：这里的想法是观察，每当N % (K + 1) = 0 的值时， A将永远无法赢得比赛。否则A总是赢得比赛。
证明：

1. If N ≤ K: The person who has the first turn will win the game, i.e. A.
2. If N = K + 1: A can remove any number of cards in the range [1, K]. So, the total number of cards left after the first turn are also in the range [1, K]. Now B gets the turn and number of cards left are in the range [1, K]. So, B will win the game.
3. If K + 2 ≤ N ≤ 2K + 1: A removes N – (K + 1) cards in his first turn. B can remove any number of cards in the range [1, K] in the next turn. Therefore, the total number of cards left now are in the range [1, K].Now, since the remaining cards left are in the range [1, K], so A can remove all the cards and win the game.

因此，我们的想法是检查N % (K + 1)是否等于 0。如果发现是真的，打印 B 作为获胜者。否则，打印 A 作为获胜者。
下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to check which
// player can win the game
void checkWinner(int N, int K)
{
    if (N % (K + 1)) {
        cout << "A";
    }
    else {
        cout << "B";
    }
}
 
// Driver code
int main()
{
 
    int N = 50;
    int K = 10;
    checkWinner(N, K);
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to check which
// player can win the game
static void checkWinner(int N, int K)
{
    if (N % (K + 1) > 0)
    {
        System.out.print("A");
    }
    else
    {
        System.out.print("B");
    }
}
 
// Driver code
public static void main(String[] args)
{
    int N = 50;
    int K = 10;
     
    checkWinner(N, K);
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement
# the above approach
 
# Function to check which
# player can win the game
def checkWinner(N, K):
 
    if(N % (K + 1)):
        print("A")
    else:
        print("B")
 
# Driver Code
N = 50
K = 10
 
# Function call
checkWinner(N, K)
 
# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to check which
// player can win the game
static void checkWinner(int N, int K)
{
    if (N % (K + 1) > 0)
    {
        Console.Write("A");
    }
    else
    {
        Console.Write("B");
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 50;
    int K = 10;
     
    checkWinner(N, K);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

    // Javascript Program to implement
// the above approach
 
// Function to check which
// player can win the game
function checkWinner(N, K)
{
    if (N % (K + 1)) {
        document.write("A");
    }
    else {
        document.write("B");
    }
}
 
// Driver code
 
    let N = 50;
    let K = 10;
    checkWinner(N, K);
 
// This code is contributed by Saurabh Jaiswal

输出：

A

时间复杂度： O(1)
辅助空间： O(1)